Find area of rectangle This is not kind of homeworks and please teke easy to consider. I found interesting problem which is very elementary but not easy (for me).
The problem is to find an area of this rectangle.
I have tried but don't know well...
Can you find it?

 A: 
Let the points starting from top left clockwisely be $A,X,B,Y,C,M,D,N$. (i.e. the topleft corner of rectangle $ABCD$ is A and topright is $B$ while the point on segment $AB$ is $X$ .etc)
Let the middle intersection be $O$. Connect $XN, YM$.
Then since $MO\cdot ON= XO\cdot OY$ by area equality and angle in between equality we know ${MO\over XO}={OY\over ON}$ and hence $XN\parallel YM$.
Hence triangles $\triangle XNA\sim\triangle MYC$ and hence ${AN\over CY}={8\over12}={2\over3}$
Also since $CY-AN=8-6=2$ we know $AN=4,CY=6$ so $AD=12$. 
Now by ratio $XO:OM=2:3$ we know $S_{XON}=20$ and $S_{YOM}=45$
Let $MD=x$ then we have $XB=x+4$
Now $S_{ABCD}=12(x+12)={8x\over2}+{6(x+4)\over2}+16+20+30+30+45+36$
Simplify we get $12x+144=7x+189\implies x=9$
Hence $S_{ABCD}=12(9+12)=252$
A: Hints:
Draw a Cartesian coordinate plane centered at the bottom left corner of the rectangle. Let the unknown length at the right side of the rectangle be $a$ and the unknown length at the bottom of the rectangle be $b$. The following is then known:

Now, you can find the coordinates of the intersection of the two lines in the middle of the rectangle. Using that point and the other two coordinates of each triangle of area 30, you can use the Shoelace Formula to set up a system of equations for $a$ and $b$. 
Edit:
Oops, I forgot the lines forming the other side of each triangle.
